Problem: A certain circle can be represented by the following equation. $x^2+y^2+10x+12y+25=0$ What is the center of this circle ? $($
Solution: The strategy We can find the center and radius of a circle by rewriting the given equation in the form of the standard equation of a circle. [What is the standard equation of the circle?] In order to do this, we take the following steps. Complete the square for both the $x^2$ and $y^2$ terms. [How do we complete the square?] Write the equation in the standard form of the circle. Completing the squares $\begin{aligned}x^2+y^2+10x+12y+25&=0\\\\ x^2+y^2+10x+12y&=-25\\\\ (x^2+10x)+(y^2+12y)&=-25 \text{(rearrange terms)}\\\\ (x^2+10x{+25})+(y^2+12y{+36})&=-25{+25}{+36}\end{aligned}$ Notice that we must add ${25}$ and ${36}$ on the right side of the equation, since we added them to the left side of the equation. [How did we get 25 and 36?] Writing the equation in standard form $\begin{aligned}(x^2+10x{+25})+(y^2+12y{+36})&=-25{+25}{+36}\\\\ (x+5)^2+(y+6)^2&=36\\\\ (x-(-5))^2+(y-(-6))^2&=6^2\end{aligned}$ Since the equation is now in the standard form, we can conclude that this circle is centered at $(-5,-6)$ and has a radius of $6$ units. Summary The circle is centered at $(-5,-6)$. The circle has a radius of $6$ units.